Question

Question #d2528

Answer 1

I get (a) 9.7 (b) 9.7 (c) 9.7

The approximation before and after the equivalence point that Le Chatelier shifts don't occur isn't great; at these low concentrations, we can't assume that concentrations on the order of `10^(-5)` `"M"` are insignificant.

Explanation:

The equation is :

`sf(HY+NaOHrarrNaY+H_2O)`

This tells us that at the equivalence point the reactants are present in the molar ratio 1 : 1.

First we find the volume of NaOH at the equivalence point:

We are told `sf(n_(HY)=0.0025)`

`:.` `sf(n_(NaOH)=0.0025)`

`sf(c=n/v)`

`:.` `sf(v=n/c=0.0025/0.1=0.025color(white)(x)L=25.00color(white)(x)ml)`

(a)

We are in the buffer region of the titration.

We need to find the pH when 24.95 ml of NaOH is added:

`sf(n_(NaOH)=cxxv=0.10xx24.95/1000=2.495xx10^(-3))`

This means that `sf(2.5xx10^(-3))` moles of NaY must form

We know that the initial moles of HY is `sf(2.50xx10^(-3))`

So the moles remaining is `sf((2.5-2.495)xx10^(-3)=5xx10^(-6))`

`sf(pK_a=7)` `:.` `sf(K_a=10^(-7))`

`sf(K_a=([H^+][Y^-])/([HY]))`

I will use moles here since the volume is constant so cancels.

`:.` `sf([H^+]=K_axx([HY])/([Y^-])=10^(-7)xx(5xx10^(-6))/(2.495xx10^(-3))`

`sf(=2.004xx10^(-10)color(white)(x)"mol/l")`

`sf(pH=-log[H^+]=-log(2.004xx10^(-10))=9.698)`

All this assumes that initial moles approximates to equilibrium moles.

(b)

At the equivalence point 0.0025 mole of HY has been converted to 0.0025 mole of NaY.

NaY is the salt of a weak acid and a strong base so undergoes hydrolysis:

`sf(Y^(-)+H_2OrightleftharpoonsHY+OH^(-))`

`sf(c=n/v)`

`:.` `sf([Y^(-)]=0.0025/((75.00+25.00)xx10^(-3))=0.025color(white)(x)"mol/l")`

`sf(pK_b=14-pK_a=14-7=7)`

`sf(pOH=1/2[pK_b-log[base]])`

`sf(pOH=1/2[7-(-1.602)]=4.301)`

`sf(pH=14-pOH=14-4.301=9.7)`

(c)

The no. moles of NaOH added is:

`sf(n_(NaOH)=cxxv=0.1xx25.05/1000=0.002505)`

The no. moles HY present = 0.0025.

This means that the no. moles XS `sf(NaOH)` is given by:

`sf(n_(XSNaOH)=0.002505-0.0025=5.00xx10^(-6))`

Total volume = 75.00 + 25.05 = 100.05 ml

`:.` `sf([NaOH]=c/v=(5.00xx10^(-6))/(100.05xx10^(-3))=0.04997xx10^(-3)color(white)(x)"mol/l")`

`sf(pOH=-log[OH^-]=-log(0.04997xx10^(-3))=4.302)`

`sf(pH=14-pOH=14-4.302~~9.698)`